一.替换所有的问号

替换所有的问号

1.思路一

class Solution {
public:string modifyString(string s) {for(int i=0;i<s.size();i++){if(s[i] == '?'){for(char j = 'a' ; j<='z' ; j++){//1.注意数组越界if((i==0 || s[i-1] != j) && (i==s.size()-1 || s[i+1] != j)){s[i] = j;break;}}}}return s;}
};

2.GIF题目解析

二.提莫攻击

提莫攻击

1.思路一

class Solution {
public:int findPoisonedDuration(vector<int>& timeSeries, int duration) {int time = 0;for(int i=1;i<timeSeries.size();i++){int time_grep = timeSeries[i] - timeSeries[i-1];//1.时间超过if(time_grep >= duration){time += duration;}//2.时间没有超过else{time += time_grep;}}//3.最后一次攻击一定会走完！time += duration;return time;}
};

三.N字型变换

1.思路一：模拟

class Solution {
public:string convert(string& s, int numRows) {//0.只有一行if (numRows == 1)return s;//1.构建二维数组vector<string> vv(numRows);//2.字符排放int flag = 1;int a = 0;int b = numRows - 2;for (int i = 0; i < s.size(); i++){//2-1:从上往下if (flag == 1){if (a <= numRows - 1){vv[a] += s[i];a++;}if (a == numRows){if(numRows >= 3)a = 0, flag = -1;a = 0;}}//2-2：从下往上else{if (b >= 0){vv[b] += s[i];b--;}if (b == 0)b = numRows-2, flag = 1;}}//3.定义字符串类型进行+=string ret;for (int i = 0; i < numRows; i++) ret += vv[i];return ret;}
};

2.思路二：找规律进行优化

class Solution{
public:string convert(string s, int numRows) {//0.特殊情况判断：if (numRows == 1)return s;//1.计算公差int d = (2 * numRows) - 2;int n = s.size();//2.规律去模拟不需要创建额外的空间：string ret;for (int i = 0; i < numRows; i++){//1.第一行字符if (i == 0)for (int i_1 = 0; i_1 < n; i_1 += d) ret += s[i_1];//2.最后一行字符else if (i == numRows - 1)for (int i_n = numRows - 1; i_n < n; i_n += d) ret += s[i_n];//3.中间字符:左在右不在特殊情况elsefor (int i_k_left = i, i_k_right = d - i; (i_k_left < n) || (i_k_right < n); i_k_left += d, i_k_right += d){if(i_k_left < n )ret += s[i_k_left];if(i_k_right < n)ret += s[i_k_right];}}return ret;}
};//时间复杂度：O(n)
//空间复杂度：O(1)

四.外观数列

外观数列

1.思路一:模拟

class Solution {
public:string countAndSay(int n) {//1.开始字符串：string ret("1");//2.内容的遍历：while (--n){int left = 0, right = 0;string s1;//3.字符串的遍历+数据更新for (left = 0, right = 0; right < ret.size(); right++){if (ret[right] != ret[left]){int n = (right - left ) * 10 + (int)(ret[left] - '0');//1.数据保存一下：s1 += to_string(n);//2.指针数值更新left = right;}}//4.数据都不去走for中的if产生的问题。int n = (right - left) * 10 + (int)(ret[left] - '0');s1 += to_string(n);ret = s1;}return ret;}
};

五.数青蛙

数青蛙

1.思路一：模拟+哈希记录数据

class Solution {
public:int minNumberOfFrogs(string croakOfFrogs) {//1.哈希表记录数据unordered_map<char, int> indx;vector<int> hash(5);//2.初始化hash的字符值？string s1("croak");//3.遍历一遍int n = s1.size();for (int i = 0; i < n; i++) indx[s1[i]] = i;//4.遍历字符串考虑青蛙！for (auto ch : croakOfFrogs){//1.第一个cif (ch == 'c'){//1.k有值if (hash[n - 1] >= 1)hash[0]++, hash[n - 1]--;//2.k没有值else if (hash[n - 1] == 0)hash[0]++;}//2.正常情况：else{int j = indx[ch];if (hash[j - 1] == 0) return -1;hash[j - 1]--, hash[j]++;}}//特殊情况：结束之后应该除了k位置hash是有数值其他hash中应该是没有数值的！for(int i=0;i<n-1;i++){if(hash[i]!=0)return -1;}return hash[n - 1];}
};